Jay2391 Posted September 20, 2006 Share Posted September 20, 2006 I hace a DB with a tabletable fields are userIDuserpassemail I am getting this error???Parse error: syntax error, unexpected $end in /home/jreina88/public_html/natgal/login.php on line 59that is the last line ...can some one help this is the codethis is the code!!!!<? session_start() ?><HTML><HEAD><TITLE>Login</TITLE></HEAD><BODY><H2>Login</H2><?php $links = "<A HREF='main.php'>Click here to proceed to the main page</A><BR><BR><A HREF='logout.php'>Click here to log out.</A>"; if($user && $pass){ if ($logged_in_user == $user) { echo $user.", you are already logged in.<BR><BR>"; echo $links; exit; } $db=mysql_connect ("localhost", "ID", "pass") or die ('I cannot connect to the database because: ' . mysql_error());mysql_select_db("database name");$result = mysql_query("SELECT * FROM tablename WHERE user = '".$user."' AND password = PASSWORD('".$pass."')"); if (!$result) { echo "Sorry, there has been a technical hitch. We cannot enter your details."; exit; } if(mysql_num_rows($result) > 0 ){ $logged_in_user = $user; session_register("logged_in_user"); echo "Welcome, ".$logged_in_user.". <br><br>"; echo $links; exit; } else { echo "Invalid login. Please try again.<br><br>"; } // else if ($user || $pass) { // echo "Please fill in both fields.<BR><BR>"; // }?><FORM METHOD=POST ACTION="login.php"> Your username: <INPUT NAME="user" TYPE=TEXT MAXLENGTH=20 SIZE=20> <BR> Your password: <INPUT NAME="pass" TYPE=PASSWORD MAXLENGTH=10 SIZE=10> <BR> <INPUT TYPE=SUBMIT VALUE="Login"> </FORM></BODY></HTML> Link to comment https://forums.phpfreaks.com/topic/21356-login-code-error/ Share on other sites More sharing options...
BillyBoB Posted September 20, 2006 Share Posted September 20, 2006 r u giving all source the error says its on 59 and theres only 49 there if i am correct Link to comment https://forums.phpfreaks.com/topic/21356-login-code-error/#findComment-95073 Share on other sites More sharing options...
Jay2391 Posted September 20, 2006 Author Share Posted September 20, 2006 I am giving you what i have .... I just do not see it...frustrating Link to comment https://forums.phpfreaks.com/topic/21356-login-code-error/#findComment-95080 Share on other sites More sharing options...
btherl Posted September 20, 2006 Share Posted September 20, 2006 Here, you open two { but only close with one } .. this could be the problem:[code] if($user && $pass){ if ($logged_in_user == $user) { echo $user.", you are already logged in.<BR><BR>"; echo $links; exit; }[/code] Link to comment https://forums.phpfreaks.com/topic/21356-login-code-error/#findComment-95083 Share on other sites More sharing options...
BillyBoB Posted September 20, 2006 Share Posted September 20, 2006 <FORM METHOD=POST ACTION="login.php"> Your username: <INPUT NAME="user" TYPE=TEXT MAXLENGTH=20 SIZE=20> <BR> Your password: <INPUT NAME="pass" TYPE=PASSWORD MAXLENGTH=10 SIZE=10> <BR> <INPUT TYPE=SUBMIT VALUE="Login"> </FORM>should be [code]<FORM METHOD="POST" ACTION="login.php"> Your username: <INPUT NAME="user" TYPE="TEXT" MAXLENGTH="20" SIZE="20"> <BR> Your password: <INPUT NAME="pass" TYPE="PASSWORD" MAXLENGTH="10" SIZE="10"> <BR> <INPUT TYPE="SUBMIT" VALUE="Login"> </FORM>[/code]just try thissorry for telling u it was only 49 i was counting them up cuz my explorer was messing up i hate it like XP well really its my comp Link to comment https://forums.phpfreaks.com/topic/21356-login-code-error/#findComment-95084 Share on other sites More sharing options...
Jay2391 Posted September 20, 2006 Author Share Posted September 20, 2006 I apreciate that i will try that tonite at home ;D Link to comment https://forums.phpfreaks.com/topic/21356-login-code-error/#findComment-95331 Share on other sites More sharing options...
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